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    Physics
    Mechanics
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    Question

    text  At one instant, the centre of mass of a system of two particles is located on the  end text x text  -axis at  end text x equals 3.0 straight m text  and has a velocity of  end text left parenthesis 6.0 straight m divided by straight s right parenthesis bold j with not stretchy hat on top text  . One of the particles is at the origin, the other particle has a  end text text  mass of  end text 0.10 kg text  and is at rest on the  end text x text  -axis at  end text x equals 12.0 straight m text  .  What is the mass of the particle at the origin?  end text

    1. 0.3 k g
    2. 0.2kg
    3. 0.5kg
    4. 0.1kg

    The correct answer is: 0.3 k g

      text  (a)  end text x subscript c M end subscript equals fraction numerator m subscript 1 x subscript 1 plus m x subscript 2 over denominator m subscript 1 plus m subscript 2 end fraction not stretchy rightwards double arrow space of 1em 3 equals fraction numerator m subscript 1 left parenthesis 0 right parenthesis plus left parenthesis 0.10 right parenthesis left parenthesis 12 right parenthesis over denominator m subscript 1 plus 0.1 end fraction
      m=0.3kg

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      text  There are two masses  end text m subscript 1 text  and  end text m subscript 2 text  placed at a distance  end text divided by text  apart. Let the centre of mass of this  end text text  system is at a point named  end text C text  . If  end text m subscript 1 text  is displaced by  end text l subscript 1 text  towards  end text C text  and  end text m subscript 2 text  is displaced by  end text I subscript 2 text  away  end text text  from  end text C text  . Find the distance, from  end text C text  where new centre of mass will be located.  end text

      text  There are two masses  end text m subscript 1 text  and  end text m subscript 2 text  placed at a distance  end text divided by text  apart. Let the centre of mass of this  end text text  system is at a point named  end text C text  . If  end text m subscript 1 text  is displaced by  end text l subscript 1 text  towards  end text C text  and  end text m subscript 2 text  is displaced by  end text I subscript 2 text  away  end text text  from  end text C text  . Find the distance, from  end text C text  where new centre of mass will be located.  end text

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      text  Two particles of masses  end text 1 kg text  and  end text 2 kg text  respectively are initially  end text 10 straight m text  apart. At time  end text t equals 0 comma text  they  end text text  start moving towards each other with uniform speeds  end text 2 straight m divided by straight s text  and  end text 1 straight m divided by straight s text  respectively. Find the  end text text  displacement of their centre of mass at  end text t equals 1 straight s text  .  end text

      text  Two particles of masses  end text 1 kg text  and  end text 2 kg text  respectively are initially  end text 10 straight m text  apart. At time  end text t equals 0 comma text  they  end text text  start moving towards each other with uniform speeds  end text 2 straight m divided by straight s text  and  end text 1 straight m divided by straight s text  respectively. Find the  end text text  displacement of their centre of mass at  end text t equals 1 straight s text  .  end text

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